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  • Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 48 maths

Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 48 maths

Answers (1)

Answer:

\frac{-1}{3 \sin ^{3} t \cos 2 t}

Hint:

You must know the derivative of cos t and sin t function

Given:

\begin{aligned} &x=3 \cos t-2 \cos ^{3} t \\ &y=3 \sin t-2 \sin ^{3} t \quad \text { find } \frac{d^{2} y}{d x^{2}} \end{aligned}

Solution:

        \begin{aligned} &x=3 \cos t-2 \cos ^{3} t \\ &\frac{d x}{d t}=-3 \sin t+6 \cos ^{2} t \sin t \\ &=\sin t+\left(6 \cos ^{2} t-3\right) \\ &y=3 \sin t-2 \sin ^{3} t \\ &\frac{d y}{d t}=3 \cos t-6 \sin ^{2} t \cos t \\ &=\cos t\left(3-6 \sin ^{2} t\right) \end{aligned}

        \begin{aligned} &\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d x}{d t}=\frac{\cos t\left(3-6 \sin ^{2} t\right)}{\sin t+\left(6 \cos ^{2} t-3\right)} \\ &=\frac{\cos t\left(1-2 \sin ^{2} t\right)}{3\left(2 \cos ^{2} t-1\right)} \\ &=\frac{\cot t(\cos 2 t)}{\cos 2 t}=\cot t \\ &\frac{d y}{d x}=\cot t \\ &\frac{d^{2} y}{d x^{2}}=-cos e c^{2} t \end{aligned}

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Gurleen Kaur

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